Variable-Frequency Response AnalysisNetwork performance as function of frequency.Transfer function

Sinusoidal Frequency AnalysisBode plots to display frequency response data

Resonant CircuitsThe resonance phenomenon and its characterization

Filter NetworksNetworks with frequency selective characteristics:low-pass, high-pass, band-pass

VARIABLE-FREQUENCY NETWORKPERFORMANCE

LEARNING GOALS

VARIABLE FREQUENCY-RESPONSE ANALYSIS

In AC steady state analysis the frequency is assumed constant (e.g., 60Hz).Here we consider the frequency as a variable and examine how the performancevaries with the frequency.

Variation in impedance of basic components

0RRZRResistor

90LLjZL Inductor

Capacitor 9011

CCjZc

Frequency dependent behavior of series RLC network

Cj

RCjLCj

CjLjRZeq

1)(1 2

C

LCjRC

j

j

)1( 2

C

LCRCZeq

222 )1()(||

RC

LCZeq

1tan

21

sC

sRCLCssZ

sj

eq1

)(2

notation" in tionSimplifica"

For all cases seen, and all cases to be studied, the impedance is of the form

011

1

011

1

...

...)(

bsbsbsb

asasasasZ n

nn

n

mm

mm

sCZsLsZRsZ CLR

1,)(,)(

Simplified notation for basic components

Moreover, if the circuit elements (L,R,C, dependent sources) are real then theexpression for any voltage or current will also be a rational function in s

LEARNING EXAMPLE

sL

sC

1

R

So VsCsLR

RsV

/1)(

SV

sRCLCs

sRC

12

So VRCjLCj

RCjV

js

1)( 2

0101)1053.215()1053.21.0()(

)1053.215(332

3

jj

jVo

LEARNING EXAMPLE A possible stereo amplifier

Desired frequency characteristic(flat between 50Hz and 15KHz)

Postulated amplifier

Log frequency scale

Frequency domain equivalent circuit

Frequency Analysis of Amplifier

)(

)(

)(

)(

sV

sV

sV

sV

in

o

S

in

)(/1

)( sVsCR

RsV S

inin

inin

]1000[/1

/1)( in

oo

oo V

RsC

sCsV

ooinin

inin

RsCRsC

RsCsG

1

1]1000[

1)(

000,40

000,40]1000[

100 ss

s

000,401001058.79

100101018.3191

1691

oo

inin

RC

RC

required

actual

000,40

000,40]1000[)(000,40||100

s

ssGs

Frequency dependent behavior iscaused by reactive elements

)(

)()(

sV

sVsG

S

o

Voltage Gain

)50( Hz

)20( kHz

NETWORK FUNCTIONS

INPUT OUTPUT TRANSFER FUNCTION SYMBOLVoltage Voltage Voltage Gain Gv(s)Current Voltage Transimpedance Z(s)Current Current Current Gain Gi(s)Voltage Current Transadmittance Y(s)

When voltages and currents are defined at different terminal pairs we define the ratios as Transfer Functions

If voltage and current are defined at the same terminals we defineDriving Point Impedance/Admittance

Some nomenclature

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gain Voltage)(

)()(

1

2

sV

sVsGv

To compute the transfer functions one must solvethe circuit. Any valid technique is acceptable

EXAMPLE

LEARNING EXAMPLE

admittanceTransfer

tanceTransadmit

)(

)()(

1

2

sV

sIsYT

gain Voltage)(

)()(

1

2

sV

sVsGv

The textbook uses mesh analysis. We willuse Thevenin’s theorem

sLRsC

sZTH ||1

)( 11

11

RsL

sLR

sC

)()(

1

112

RsLsC

RsLLCRssZTH

)()( 11

sVRsL

sLsVOC

)(sVOC

)(sZTH

)(2 sV

2R)(2 sI

)(

)()(

22 sZR

sVsI

TH

OC

)(

)(

1

112

2

11

RsLsCRsLLCRs

R

sVRsL

sL

121212

2

)()()(

RCRRLsLCRRs

LCssYT

2 2 22

1 1

( ) ( )( ) ( )

( ) ( )v T

V s R I sG s R Y s

V s V s

)(

)(

1

1

RsLsC

RsLsC

POLES AND ZEROS(More nomenclature)

011

1

011

1

...

...)(

bsbsbsb

asasasasH n

nn

n

mm

mm

Arbitrary network function

Using the roots, every (monic) polynomial can be expressed as aproduct of first order terms

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

function network the of poles

function network the of zeros

n

m

ppp

zzz

,...,,

,...,,

21

21

The network function is uniquely determined by its poles and zerosand its value at some other value of s (to compute the gain)

EXAMPLE

1)0(

22,22

,1

21

1

H

jpjp

z

:poles

:zeros

)22)(22(

)1()( 0 jsjs

sKsH

84

120

ss

sK

18

1)0( 0KH

84

18)( 2

ss

ssH

LEARNING EXTENSION

)104(

000,20,50

0

70

21

1

K

HzpHzp

z

:poles

:zero

)(

)()(

sV

sVsG

K

S

o

o

gain voltagethefor

of valuethe and locations zero and pole the Find

ooinin

inin

RsCRsC

RsCsG

1

1]1000[

1)(

000,40

000,40]1000[

100 ss

sFor this case the gain was shown to be

))...()((

))...()(()(

21

210

n

m

pspsps

zszszsKsH

Zeros = roots of numeratorPoles = roots of denominator

VariableFrequencyResponse

SINUSOIDAL FREQUENCY ANALYSIS

)(sH

Circuit represented bynetwork function

)cos(0

)(0

tB

eA tj

)(cos|)(|

)(

0

)(0

jHtjHB

ejHA tj

)()()(

)()(

|)(|)(

jeMjH

jH

jHM

Notation

stics.characteri phase and magnitude

calledgenerally are of function as of Plots ),(),(M

)(log)(

))(log2010

10

PLOTS BODE vs

(M

. of function a as function network the

analyze wefrequency the of function a as network a ofbehavior thestudy To

)( jH

HISTORY OF THE DECIBEL

Originated as a measure of relative (radio) power

1

2)2 log10(|

P

PP dB 1Pover

21

22

21

22)2

22 log10log10(|

I

I

V

VP

R

VRIP dB 1Pover

By extension

||log20|

||log20|

||log20|

10

10

10

GG

II

VV

dB

dB

dB

Using log scales the frequency characteristics of network functionshave simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations

]...)()(21)[1(

]...)()(21)[1()()( 2

233310

bbba

N

jjj

jjjjKjH

General form of a network function showing basic terms

Frequency independent

Poles/zeros at the origin

First order terms Quadratic terms for complex conjugate poles/zeros

..|)()(21|log20|1|log20

...|)()(21|log20|1|log20

||log20log20

21010

233310110

10010

bbba jjj

jjj

jNK

DND

N

BAAB

loglog)log(

loglog)log(

|)(|log20|)(| 10 jHjH dB

212

1

2121

zzz

z

zzzz

...)(1

2tantan

...)(1

2tantan

900)(

211

23

3311

1

b

bba

NjH

Display each basic termseparately and add theresults to obtain final answer

Let’s examine each basic term

Constant Term

Poles/Zeros at the origin

90)(

)(log20|)(|)( 10

Nj

Njj

NdB

NN

linestraight a is this

log is axis- xthe 10

Simple pole or zero j1

1

210

tan)1(

)(1log20|1|

j

j dB

asymptotefrequency low 0|1| dBj

(20dB/dec) asymptotefrequency high 10log20|1| dBj

frequency) akcorner/bre1 whenmeet asymptotes two The (

Behavior in the neighborhood of the corner

FrequencyAsymptoteCurvedistance to asymptote Argument

corner 0dB 3dB 3 45

octave above 6dB 7db 1 63.4

octave below 0dB 1dB 1 26.6

125.0

0)1( j

90)1( j

1

1

Asymptote for phase

High freq. asymptoteLow freq. Asym.

Simple zero

Simple pole

Quadratic pole or zero ])()(21[ 22 jjt ])()(21[ 2 j

222102 2)(1log20|| dBt 2

12

)(1

2tan

t

1 asymptotefrequency low 0|| 2 dBt 02t

1 asymptote freq. high 2102 )(log20|| dBt 1802t1 )2(log20|| 102 dBt 902tCorner/break frequency

221 2102 12log20|| dBt

21

221

tan

t2

2

Magnitude for quadratic pole Phase for quadratic pole

dB/dec40

These graphs are inverted for a zero

LEARNING EXAMPLEGenerate magnitude and phase plots

)102.0)(1(

)11.0(10)(

jj

jjGvDraw asymptotes

for each term1,10,50 :nersBreaks/cor

40

20

0

20

dB

90

90

1.0 1 10 100 1000

dB|10

decdB /20

dec/45

decdB /20

dec/45

Draw composites

asymptotes

LEARNING EXAMPLEGenerate magnitude and phase plots

)11.0()(

)1(25)( 2

jj

jjGv 10 1, :(corners) Breaks

40

20

0

20

dB

90

270

90

1.0 1 10 100

Draw asymptotes for each

dB28

decdB /40

180

dec/45

45

Form composites

21

020 0)(

Kj

K

dB

Final results . . . And an extra hint on poles at the origin

dec

dB40

dec

dB20

dec

dB40

LEARNING EXTENSION Sketch the magnitude characteristic

)100)(10(

)2(10)(

4

jj

jjG

form standard in NOT is function theBut

100 10, 2, :breaks

Put in standard form)1100/)(110/(

)12/(20)(

jj

jjG We need to show about

4 decades

40

20

0

20

dB

90

90

1 10 100 1000

26 |dB

LEARNING EXTENSION Sketch the magnitude characteristic

2)(

102.0(100)(

j

jjG

origin theat pole Double

50at break

form standard in isIt

40

20

0

20

dB

90

270

90

1 10 100 1000

Once each term is drawn we form the composites

Put in standard form

)110/)(1()(

jj

jjG

LEARNING EXTENSION Sketch the magnitude characteristic

)10)(1(

10)(

jj

jjG

10 1, :breaks

origin theat zero

form standard innot

40

20

0

20

dB

90

270

90

1.0 110

100Once each term is drawn we form the composites

decdB /20decdB /20

DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT

This is the inverse problem of determining frequency characteristics. We will use only the composite asymptotes plot of the magnitude to postulatea transfer function. The slopes will provide information on the order

A

A. different from 0dB.There is a constant Ko

B

B. Simple pole at 0.11)11.0/( j

C

C. Simple zero at 0.5

)15.0/( j

D

D. Simple pole at 3

1)13/( j

E

E. Simple pole at 20

1)120/( j

)120/)(13/)(11.0/(

)15.0/(10)(

jjj

jjG

20

|

00

0

1020|dBK

dB KK

If the slope is -40dB we assume double real pole. Unless we are given more data

LEARNING EXTENSIONDetermine a transfer function from the composite magnitude asymptotes plot

A

A. Pole at the origin. Crosses 0dB line at 5

j5

B

B. Zero at 5

C

C. Pole at 20

D

D. Zero at 50

E

E. Pole at 100

)1100/)(120/(

)150/)(15/(5)(

jjj

jjjG

Sinusoidal

RESONANT CIRCUITS

These are circuits with very special frequency characteristics…And resonance is a very important physical phenomenon

LCCL

110

when zero iscircuit each of reactance The

The frequency at which the circuit becomes purely resistive is calledthe resonance frequency

CjLjRjZ

1

)(

circuit RLC Series

LjCjGjY

1

)(

circuit RLC Parallel

Properties of resonant circuits

At resonance the impedance/admittance is minimal

Current through the serial circuit/voltage across the parallel circuit canbecome very large

CRR

LQ

0

0 1

:FactorQuality

222 )1

(||

1)(

CLRZ

CjLjRjZ

222 )1

(||

1)(

LCGY

CjLj

GjY

Given the similarities between series and parallel resonant circuits, we will focus on serial circuits

Properties of resonant circuits

CIRCUIT BELOW RESONANCE ABOVE RESONANCESERIES CAPACITIVE INDUCTIVEPARALLEL INDUCTIVE CAPACITIVE

Phasor diagram for series circuit Phasor diagram for parallel circuit

RV

C

IjVC

Lj

1GV 1CVj

L

Vj1

LEARNING EXAMPLEDetermine the resonant frequency, the voltage across eachelement at resonance and the value of the quality factor

LC

10 sec/2000

)1010)(1025(

163

radFH

I

AZ

VI S 5

2

010

2Z resonanceAt

50)1025)(102( 330L

)(902505500 VjLIjVL

902505501

501

0

00

jICj

V

LC

C

R

LQ 0 25

2

50

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

LEARNING EXTENSION Find the value of C that will place the circuit in resonanceat 1800rad/sec

LC

10 218001.0

1

)(1.0

11800

C

CH

FC 86.3

Find the Q for the network and the magnitude of the voltage across thecapacitor

R

LQ 0 60

3

1.01800

Q

||||

|||| 0

SC

SS

L

VQV

VQR

VLV

resonanceAt

VVC 600||

Resonance for the series circuit

222 )1

(||

1)(

CLRZ

CjLjRjZ

QRCQRL

1, 00

:resonanceAt

)(1

)(

0

0

0

0

jQR

QRjQRjRjZ

)(1

1

0

0

1

jQV

VG Rv

is gain voltageThe :Claim

)(1

jZ

R

CjLjR

RGv

vv GGM |)(|,|)(

2/120

0

2 )(1

1)(

Q

M

)(tan)( 0

0

1

Q

QBW 0

12

1

2

12

0 QQLO

sfrequenciepower Half

Z

RGv

LEARNING EXAMPLE

2

mH2

F5

Determine the resonant frequency, quality factor andbandwidth when R=2 and when R=0.2

CRR

LQ

0

0 1

LC

10

QBW 0

sec/10)105)(102(

1 4

630 rad

R Q2 10

0.2 100

R Q BW(rad/sec)2 10 1000

0.2 100 100

LEARNING EXTENSIONA series RLC circuit as the following properties:sec/100sec,/4000,4 0 radBWradR

Determine the values of L,C.

CRR

LQ

0

0 1

LC

10

QBW 0

1. Given resonant frequency and bandwidth determine Q.2. Given R, resonant frequency and Q determine L, C.

40100

40000 BW

Q

HQR

L 040.04000

440

0

FRQL

C 662

020

1056.11016104

111

PROPERTIES OF RESONANT CIRCUITS: VOLTAGE ACROSS CAPACITOR

|||| 0 RVQV resonanceAt

But this is NOT the maximum value for thevoltage across the capacitor

CRjLCCj

LjR

CjV

V

S

20

1

11

1

2

221

1)(

Qu

u

ug

2

0

0

;SV

Vgu

22

22

2

1

)/1)(/(2)2)(1(20

Qu

u

QQuuu

du

dg

CRR

LQ

0

0 1

LC

10

22 1)1(2Q

u

20

maxmax

2

11

Qu

2

2

424

max

4

11

2

11

4

1

1

Q

Q

QQQ

g

2

0

4

11

||||

Q

VQV S

LEARNING EXAMPLE

mH50

F5

150, RR and when Determine max0

sradLC

/2000)105)(105(

11620

CRR

LQ

0

0 1

LC

10

20

maxmax

2

11

Qu

RQ

050.02000 2max 2

112000Q

R Q Wmax50 2 18711 100 2000

Evaluated with EXCEL and rounded to zero decimals

Using MATLAB one can display the frequency response

R=50Low QPoor selectivity

R=1High QGood selectivity

FILTER NETWORKS

Networks designed to have frequency selective behavior

COMMON FILTERS

Low-pass filterHigh-pass filter

Band-pass filter

Band-reject filter

We focus first onPASSIVE filters

Simple low-pass filter

RCjCj

R

CjV

VGv

1

11

1

1

0

RCj

Gv

;

1

1

1

2

tan)(

1

1||)(

v

v

G

GM

2

11,1max

MM

frequencypower half

1

1

BW

Simple high-pass filter

CRj

CRj

CjR

R

V

VGv

11

1

0

RCj

jGv

;1

1

2

tan2

)(

1||)(

v

v

G

GM

2

11,1max

MM

frequencypower half

1

1

LO

Simple band-pass filter

Band-pass

CLjR

R

V

VGv

11

0

222 1)(

LCRC

RCM

11

LCM 0)()0( MM

2

4/)/( 20

2

LRLRLO

LC

10

2

4/)/( 20

2

LRLRHI

L

RBW LOHI

)(2

1)( HILO MM

Simple band-reject filter

011

000

C

LjLC

10 VV circuit open as actscapacitor the 0at

10 VV circuit open as actsinductor the at

filter pass-band

the in as determined are HILO ,

LEARNING EXAMPLEDepending on where the output is taken, this circuitcan produce low-pass, high-pass or band-pass or band-reject filters

Band-pass

Band-reject filter

CLjR

Lj

V

V

S

L

1 1)(,00

S

L

S

L

V

V

V

VHigh-pass

CLjR

CjV

V

S

C

1

1

0)(,10 S

C

S

C

V

V

V

VLow-pass

FCHLR 159,159,10 for plot Bode

ACTIVE FILTERS

Passive filters have several limitations

1. Cannot generate gains greater than one

2. Loading effect makes them difficult to interconnect

3. Use of inductance makes them difficult to handle

Using operational amplifiers one can design all basic filters, and more, with only resistors and capacitors

The linear models developed for operational amplifiers circuits are valid, in amore general framework, if one replaces the resistors by impedances

Ideal Op-Amp

These currents arezero

Basic Inverting Amplifier

0V

VV gain Infinite

0V

0 II -impedanceinput Infinite

1

1 2

0OVV

Z Z

21

1O

ZV V

Z

Linear circuit equivalent

0I

1

2

Z

ZG

1

11 Z

VI

EXAMPLE USING INVERTING AMPLIFIER

LOW PASS FILTER

Basic Non-inverting amplifier

1V

1V

0I

1

1

2

10

Z

V

Z

VV

11

120 V

Z

ZZV

1

21Z

ZG

01 I

EXAMPLEUSING NON INVERTING CONFIGURATION

EXAMPLE SECON ORDER FILTER

2( )V s

22 2 2

1 1 2 3

01/

OINV VV V V V

R C s R R

2

2 2

01/

OVV

R C s